Sine/ Cosine Rule Problems

**ns2583** · July 5th 2009, 01:16 AM

Hi. So, I'm having problems figuring out this problem. For a, the furthest I got up to was $x^2 -(14)(cos50)x -51=0$ and i'm not even sure that's right. Some help would be appreciated.

Thanks!

**Prove It** · July 5th 2009, 01:26 AM

Originally Posted by ns2583

Hi. So, I'm having problems figuring out this problem. For a, the furthest I got up to was $x^2 -(14)(cos50)x -51=0$ and i'm not even sure that's right. Some help would be appreciated.

Thanks!

Your equation is correct.

You can solve it for $x$ using the Quadratic formula.

**mr fantastic** · July 5th 2009, 01:28 AM

Originally Posted by ns2583

Hi. So, I'm having problems figuring out this problem. For a, the furthest I got up to was $x^2 -(14)(cos50)x -51=0$ and i'm not even sure that's right. Some help would be appreciated.

Thanks!

(a) is a direct application of the cosine rule: $10^2 = 7^2 + x^2 - 2 (x)(7) \cos 50^0$ .

(b) $\frac{x}{\sin \alpha} = \frac{10}{\sin 50^0}$ .... (1)

$\frac{x}{\sin \alpha} = \frac{10}{\sin 50^0}$ .... (1)

$\frac{7}{\sin \beta} = \frac{10}{\sin 50^0}$ .... (2)

From (2) get $\beta$ . From $\alpha + \beta + 50^0 = 180^0$ get $\alpha$ . Now use (1) to solve for $x$ .

**ns2583** · July 5th 2009, 08:21 AM

Originally Posted by Prove It

Your equation is correct.

You can solve it for $x$ using the Quadratic formula.

When I put it in my calc, it said there was a nonreal answer. :S

Originally Posted by mr fantastic

(a) is a direct application of the cosine rule:

(b) $\frac{x}{\sin \alpha} = \frac{10}{\sin 50^0}$ .... (1)

$\frac{x}{\sin \alpha} = \frac{10}{\sin 50^0}$ .... (1)

$\frac{7}{\sin \beta} = \frac{10}{\sin 50^0}$ .... (2)

From (2) get $\beta$ . From $\alpha + \beta + 50^0 = 180^0$ get $\alpha$ . Now use (1) to solve for $x$ .

Got it, thanks!

**yeongil** · July 5th 2009, 02:19 PM

Originally Posted by ns2583

Originally Posted by Prove It

Your equation is correct.

You can solve it for $x$ using the Quadratic formula.

When I put it in my calc, it said there was a nonreal answer. :S

Huh?

Just use the quadratic formula:

$\begin{aligned} x^2 - 14(\cos 50^{\circ}) x - 51 &= 0 \\ x^2 - 9x - 51 &\approx 0 \\ x &\approx \frac{9 \pm \sqrt{(-9)^2 - 4(1)(-51)}}{2(1)} \\ &\approx \frac{9 \pm \sqrt{284.98}}{2} \\ \end{aligned}$

$x \approx 12.94$
$x \approx -3.94$
Reject the 2nd solution, so x is about 13 cm.

01

**ns2583** · July 5th 2009, 06:42 PM

OH, i see. Thanks.

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